Given, the maximum electron density in the morning

$N_{max}={10}^{10} {\mathrm{m}}^{-3}$

$f_c^{\text{'}}=9{\left(N_{max}\right)}^{1/2}=9{\left({10}^{10}\right)}^{1/2}$

Critical frequency $f_c^{\prime }=9\times {10}^5$

The maximum electron density at noon

$N_{max}={2\times 10}^{10} {\mathrm{m}}^{-3}$

Critical frequency

$f_c^{\prime \prime }=9\times {\left(N_{max}^{\prime }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=9\times {\left(2\times {10}^{10}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=9\times \sqrt{2}\times {10}^5$

The ratio of critical frequency at noon and critical frequency in the morning.

$\frac{f_c^{\prime \prime }}{f_c^{\prime }}=\frac{9\times \sqrt{2}\times {10}^5}{9\times {10}^5}$

Or $\frac{f_c^{\prime \prime }}{f_c^{\prime }}=\sqrt{2}=1.414$